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using F'(a)=lim h->0 (f(a+h)-f(a))/h f(x)=1/x,a=2
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\[\lim_{h \rightarrow 0}\frac{ \frac{ 1 }{ x+h } -\frac{ 1 }{ x }}{ h }\]
i was able to get that far, the answer is -1/4 i am not sure what to do after that
\[\lim_{h \rightarrow0}\frac{ \frac{ x }{ x(x+h) } -\frac{ (x+h) }{ x(x+h) }}{ h } \rightarrow \lim_{h \rightarrow 0} \frac{ \frac{ -h }{ x(x+h) } }{ h } \rightarrow \lim_{h \rightarrow 0} \frac{ -1 }{ x(x+h) }\]
thank you!
this should help you to solve by plugging in 2 for x
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